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Simple Interest examples ( Enter your problem )
( Enter your problem )
 
  1. Example-1
 		Other related methods
  1. Statistics Word Problem
  2. HCF-LCM Word Problem
  3. Percentage
  4. Profit Loss Discount
  5. Simple Interest
  6. Compound Interest
  7. Installment
  8. Arithmetic Progression
  9. Geometric Progression
4. Profit Loss Discount
(Previous method)		6. Compound Interest
(Next method)

1. Example-1
Problem : 1 / 13 [ Simple Interest ]       Enter your problem
1. Find the simple interest on Rs 730 for 184 days at 25/4 % per annum.
Solution: Here `P = `Rs `730, R = 25/4` % and Time `= 184` days = `184/365` years

`SI = (P*R*N)/100 = ( 730 * 25/4 * 184 /365) / 100 = 23`

Simple Interest is Rs `23` .

Problem : 2 / 13 [ Simple Interest ]       Enter your problem
2. Find the simple interest on Rs 4660 for 5 years at 7/2 % per annum.
Solution: Here `P = ` Rs `4660, N = 5` years and `R = 7/2` %

`SI = (P*R*N)/100 = ( 4660 * 7/2 * 5 ) / 100 = 815.5`

Simple Interest is Rs `815.5` .

Problem : 3 / 13 [ Simple Interest ]       Enter your problem
3. The interest on a certain amount of money at 8 % per year for a period of 4 years is Rs 512 . Find the sum of money.
Solution: Here `SI = `Rs `512, N = 4` years, `R = 8` %

`SI = (P*R*N)/100`

`P = (SI*100) / (R*N)`

`P = (512 * 100 ) / ( 8 * 4 ) = 1600`

The sum of money is Rs `1600` .

Problem : 4 / 13 [ Simple Interest ]       Enter your problem
4. Simple Interest on Rs 972 at 14 % per annum for a certain time is Rs 476.28 . Find the time ?
Solution: Here `P = 972, SI = `Rs `476.28, R = 14`%

`SI = (P*R*N)/100 `

`N = (SI*100) / (P*R)`

`N = ( 476.28 * 100 ) / ( 972 * 14 ) = 3.5`

The time is `3.5` years

Problem : 5 / 13 [ Simple Interest ]       Enter your problem
5. Simple Interest on Rs 972 at a certain rate % per annum for 3.5 years is Rs 476.28 . Find the rate ?
Solution: Here `P = 972, SI = ` Rs `476.28, N = 3.5`

`SI = (P*R*N)/100`

`R = (SI*100) / (P*N) = ( 476.28 * 100 ) / ( 972 * 3.5 ) = 14`

The rate is `14` %.

Problem : 6 / 13 [ Simple Interest ]       Enter your problem
6. A sum of money lent at simple interest amounts to Rs 1008 in 2 years and to 1112 in 3 years.Find the sum and the rate of interest.
Solution: Principal + Interest for `3` years = `1112`

Principal + Interest for `2` years = `1008`

`=>` Interest for `1` year = `1112 - 1008 = 104`

`=>` Interest for `2` years = `2 * 104 = 208`

Now, Principal + Interest for `2` years = `1008`

`=>` Principal + `208 = 1008`

`=>` Principal = `1008 - 208 = 800`

`SI = (P*R*N)/100`

`R = (SI*100) / (P*N) = ( 208 * 100 ) / ( 800 * 2 ) = 13`

The Sum of money is Rs `800` and the rate of interest is `13` %

Problem : 7 / 13 [ Simple Interest ]       Enter your problem
7. What sum of money lent out at simple interest at 9 % p.a. for 3/2 years will produce the same interest as Rs 2250 lent at 6 % p.a. for 5 years.
Solution: Here `P = ` Rs `2250, N = 5` years, `R = 6`%

`SI = (P*R*N)/100`

`SI = (2250 * 6 * 5 ) / 100 = 675`

Now, `SI = ` Rs `675, N = 3/2` years, `R = 9`%

`SI = (P*R*N)/100`

`P = (SI * 100) / (R * N) = ( 675 * 100) / ( 9 * 3/2 ) = 5000`

The sum of money is Rs `5000`


Problem : 8 / 13 [ Simple Interest ]       Enter your problem
8. A man puts out Rs 500 for 4 years on simple interest and Rs 600 for 3 years. The total interest he receives is Rs 190 . What is the rate percent per annum?
Solution: Let the rate(`R`) % per annum = `X`

`SI` on Rs `500 = (P*R*N)/100 = (500 * X * 4) / 100 = 20 X`

`SI` on Rs `600 = (P*R*N)/100 = (600 * X * 3) / 100 = 18 X`

Total `SI` = Rs `190`

`20 X + 18 X = 190`

`38 X = 190`

`X = 190 / 38 = 5`

Rate is `5` %

Problem : 9 / 13 [ Simple Interest ]       Enter your problem
9. A sum was put at simple interest at a certain rate for 2 years. Had it been put at 3 % higher rate, it whould have fetched Rs 300 more. Find the sum.
Solution: Let sum = Rs `X` and original rate = `R` %

Then,
`(X * ( R + 3 ) * 2)/100 - (X * R * 2)/100 = 300`

`2 R X + 6 X - 2 R X = 300 * 100`

`6 X = 300 * 100`

`X = 5000`

The sum is Rs `5000` .

Problem : 10 / 13 [ Simple Interest ]       Enter your problem
10. Divide Rs 16875 into two parts such that the interest at 7/2 % p.a. for 2 years on one part is equal to the other at 4 % p.a. for 5 years.
Solution: Let the Principal of one part = `X` Rs

The Principal of second part = `( 16875 - X )` Rs

For one part, `P = X` Rs, `R = 7/2 %, N = 2` years

`SI = (P*R*N)/100`

`SI = ( X * 7/2 * 2 ) / 100 = (7 * X)/100 ->(1)`

Now for second part, `P = ( 16875 - X ) Rs, R = 4 %, N = 5` years

`SI = (P*R*N)/100`

`SI = (( 16875 - X ) * 4 * 5 ) / 100 = (20 * ( 16875 - X ))/100 ->(2)`

Interest is the same as `(1) = (2)`

`=> (7 * X)/100 = (20 * ( 16875 - X ))/100`

`=> 7 X = 20 * ( 16875 - X )`

`=> 7 X = 20 * 16875 - 20 X`

`=> 27 X = 20 * 16875`

`=> X = (20 * 16875) / 27`

`=> X = 12500`

`:.`One part of Principal = Rs `12500` and the second part = Rs `( 16875 - 12500 )` = Rs `4375`

The two parts are Rs `12500` and Rs `4375` .

Problem : 11 / 13 [ Simple Interest ]       Enter your problem
11. A shopkeeper borrowed Rs 20000 from two money lendeRs For one loan he paid 12 % and for the other 14 % per annum. After 1 year, he paid Rs 2560 as interest. How much did he borrow at each rate ?
Solution: Let money at `12 % = X` and that at `14 % = ( 20000 - X )`

After one year total interest = `2560`

`(X * 12 * 1) / 100 + ((20000 - X) * 14 * 1) / 100 = 2560`

`12 X + 14 * 20000 - 14 X = 2560 * 100`

`-2 X = -24000`

`X = 12000`

`:.`Money at `12` % = Rs `12000` and Money at `14` % = Rs `8000` .

Problem : 12 / 13 [ Simple Interest ]       Enter your problem
12. At what rate percent per annum will sum of money double in 8 years?
Solution: Let, Sum = `X`, Time = `8` years and Amount = `2X`.

`:. SI = X. `

`SI = (P*R*N)/100`

`R = (SI * 100)/(P * N) = (X * 100)/( X * 8 ) = 25/2` %

`:.` sum of money double at `25/2` % in `8` years.

Problem : 13 / 13 [ Simple Interest ]       Enter your problem
13. Rajeev deposited money in the post office which is doubled in 20 years at a simple rate of interest. In how many years will the original sum triple itself ?
Solution: Let, Sum = `X`, Time = `20` years and Amount = `2X`.

`:. SI = X. `

`SI = (P*R*N)/100`

`R = (SI * 100)/(P * N) = (X * 100)/( X * 20 ) = 5` %

`:.` sum of money double at `5` % in `20` years.

4. Profit Loss Discount
(Previous method)		6. Compound Interest
(Next method)


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